# HOW TO CHECK IF A GIVEN POINT LIES INSIDE OR OUTSIDE A CIRCLE

## About "How to Check if a Given Point Lies Inside or Outside a Circle"

How to Check if a Given Point Lies Inside or Outside a Circle :

Here we are going to see, how to check if a given point lies inside or outside a circle.

x12 + y12 + 2gx1 + 2fy1 + c  >  0 (outside of the circle)

x12 + y12 + 2gx1 + 2fy1 + c  =  0 (on the circle)

x12 + y12 + 2gx1 + 2fy1 + c  <  0 (inside the circle)

## How to Check if a Given Point Lies Inside or Outside a Circle - Examples

Question 1 :

Determine whether the points (-2, 1),(0, 0) and (-4, -3) lie outside, on or inside the circle x2 + y2 − 5x + 2y − 5 = 0 .

Solution :

Equation of the circle :

x2 + y2 − 5x + 2y − 5

Equation of the circle at the point (x1, y1) :

x12 + y12 − 5x1 + 2y1 − 5

at (-2, 1)

=  (-2)2 + 12 − 5(-2) + 2(1) − 5

=  5 + 10 + 2 - 5

=  12 > 0

So, the point (-2, 1) lies outside the circle.

at (0, 0)

=  02 + 02 − 5(0) + 2(0) − 5

=  - 5 < 0

So, the point (0, 0) lies inside the circle.

at (-4, -3)

=  (-4)2 + (-3)2 − 5(-4) + 2(-3) − 5

=  16 + 9 + 20  - 6 - 5

=  34 > 0

So, the point (-4, -3) lies outside the circle.

## Finding Center and Radius of the Circle

Question 2 :

Find centre and radius of the following circles.

(i) x2 + (y + 2)2 = 0

Solution :

By comparing the given equation with general form of a circle, we get

x2 + (y + 2)2 = 0

(x - h)2 + (y - k)2 = r2

(h, k) ==>  (0, -2) and r is 0.

Hence the center and radius of the circle are (0, -2) and 0.

(ii)  x2 + y2 + 6x − 4y + 4 = 0

Solution :

Center of the circle is (-g, -f)

radius of the circle  =  √g2 + f2 - c

g = 3, f = -2 and c = 4

=  √32 + (-2)2 - 4

=  √(9 + 4 - 4)

=  √9

=  3

Hence the center and radius of the circle are (-3, 2) and 3 respectively.

(iii) x2 + y2 − x + 2y − 3 = 0

Solution :

g = -1/2, f = 1 and c = -3

=  √(-1/2)2 + 12 + 3

=  √((1/4) + 4)

=  √17/4

=  √17/2

Hence the center and radius of the circle are (1/2, -1) and √17/2 respectively.

(iv) 2x2 + 2y2 − 6x + 4y + 2 = 0

Solution :

By dividing the entire equation by 2, we get

x2 + y2 − 3x + 2y + 1 = 0

g = -3/2, f = 1 and c = 1

=  √(-3/2)2 + 12 - 1

=  √(9/4)

=  3/2

Hence the center and radius of the circle are (3/2, -1) and 3/2 respectively.

Question 3 :

If the equation 3x2 + (3 − p) xy + qy2 − 2 px = 8pq represents a circle, find p and q . Also determine the centre and radius of the circle.

Solution :

The equation x2 + y2 + 2gx + 2fy + c = 0 is a second degree equation in x and y possessing the following characteristics:

(i) It is a second degree equation in x and y ,

(ii) coefficient of x2 = coefficient of y≠ 0,

(iii) coefficient of xy = 0 .

3x2 + (3 − p) xy + qy2 − 2 px = 8pq

According to (iii), coefficient of x y  = 0

3 - p  =  0

p  =  3

According to (ii), coefficient of x2 = coefficient of y≠ 0

coefficient of x2 = 3 = coefficient of y

q = 3

By applying the values of p and q in the given equation, we get

3x2 + 3y2 − 6x = 72

x2 + y2 − 2x = 24

Center of the circle (-g, -f)

g = -1, f = 0 and c = -24

Center of the circle  =  (1, 0)

Radius of the circle  =  √g2 + f2 - c

=  √(-1)2 + 02 + 24

=  √25  =  5

Hence the center and radius of the circle are (1, 0) and 5.

After having gone through the stuff given above, we hope that the students would have understood, "How to Check if a Given Point Lies Inside or Outside a Circle".

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